MHB What is the Explanation for the Intermediate Value Theorem 2 Proof?

[solakis1]

If:

f is a real value function continuous over [a,b] ,a<b and f(a)<f(b)

,then prove that for all η,
f(a)<η<f(b) ,there exists an ξ ,such that : f(ξ)=η

Proof.

1)Let S be a set of all xε[a,b] such that :f(x)<η.

Since a belongs to S and b is an upper bound then S has a Supremum

2)Let ξ= SupS3) Will show that : a) $f(\xi)\geq\eta $ ,b) $ f(\xi)\leq\eta $ and hence : $f(\xi)=\eta$3a) By the definition of supremum, for every positive ε, there is a number x' of S with
$\xi-\epsilon\leq x'\leq\xi$ .
For this x', f(x')<η.
Since f is contiguous at ξ, $f(\xi)\leq\eta$ which is : 3a

3b) Any x greater than ξ is not in S and so $f(x)\geq \eta$.By continuity ,f(ξ) is the limit of f(x) as x tends to ξ through values greater than ξ ,and so :$f(\xi)\geq\eta$

I have no idea how he gets $f(\xi)\geq \eta $ in part 3b and $f(\xi)\leq\eta$ in part 3a

 

[Amer]

Re: intermediate value theorem 2

If:

f is a real value function continuous over [a,b] ,a<b and f(a)<f(b)

,then prove that for all η,
f(a)<η<f(b) ,there exists an ξ ,such that : f(ξ)=η

Proof.

1)Let S be a set of all xε[a,b] such that :f(x)<η.

Since a belongs to S and b is an upper bound then S has a Supremum

2)Let ξ= SupS3) Will show that : a) $f(\xi)\geq\eta $ ,b) $ f(\xi)\leq\eta $ and hence : $f(\xi)=\eta$3a) By the definition of supremum, for every positive ε, there is a number x' of S with
$\xi-\epsilon\leq x'\leq\xi$ .
For this x', f(x')<η.
Since f is contiguous at ξ, $f(\xi)\leq\eta$ which is : 3a

3b) Any x greater than ξ is not in S and so $f(x)\geq \eta$.By continuity ,f(ξ) is the limit of f(x) as x tends to ξ through values greater than ξ ,and so :$f(\xi)\geq\eta$

I have no idea how he gets $f(\xi)\geq \eta $ in part 3b and $f(\xi)\leq\eta$ in part 3a

3a) f(\xi) \leq \eta
Pick \epsilon = \eta - f(\xi) as you should know
f is continuous at ξ that means for any \epsilon &gt;0 there exist \delta &gt;0
such that whenever \mid x - \xi \mid &lt; \delta we have \mid f(x) - f(\xi) \mid &lt; \epsilon
remember that epsilon>0 so if you choose a wrong epsilon the criteria for the continuity will be affected, back to our proof after we choose epsilon sub it
\mid f(x) - f(\xi) \mid &lt; \eta - f(\xi)

-\eta + f(\xi) &lt; f(x) - f(\xi) &lt; \eta - f(\xi)

f(x) - f(\xi) &lt; \eta - f(\xi) \Rightarrow f(x) &lt; \eta for

\mid x - \xi \mid &lt; \delta \Rightarrow \xi - \delta &lt; x &lt; \xi + \delta

but \xi is an upper bound for S, \xi &lt; \xi +\frac{\delta}{2}
and
\xi + \frac{\delta}{2} \in S since
f\left(\xi + \frac{\delta}{2}\right) &lt; \eta but \xi is an upper bound for S contradiction! so \eta - f(\xi) is not positive so still two choices less than or equal zero

Lets see if f(\xi) &gt; \eta
f is continuous at \xi , now pick \epsilon = f(\xi) - \eta
there exist delta

\mid f(x) - f(\xi) \mid &lt; f(\xi) - \eta for \mid x - \xi \mid &lt; \delta

-f(\xi) + \eta &lt; f(x) - f(\xi)

\eta &lt; f(x) for \xi - \delta &lt; x &lt; \xi + \delta

\xi - \delta is in S the definition of the upper bound which means f \left(\xi - \delta \right) should be less than \eta not larger contradiction so our epsilon is not positive

This idea from Wiki see this

 

[solakis1]

Re: intermediate value theorem 2

3a) f(\xi) \leq \eta
Pick \epsilon = \eta - f(\xi) as you should know
f is continuous at ξ that means for any \epsilon &gt;0 there exist \delta &gt;0
such that whenever \mid x - \xi \mid &lt; \delta we have \mid f(x) - f(\xi) \mid &lt; \epsilon
remember that epsilon>0 so if you choose a wrong epsilon the criteria for the continuity will be affected, back to our proof after we choose epsilon sub it
\mid f(x) - f(\xi) \mid &lt; \eta - f(\xi)

-\eta + f(\xi) &lt; f(x) - f(\xi) &lt; \eta - f(\xi)

f(x) - f(\xi) &lt; \eta - f(\xi) \Rightarrow f(x) &lt; \eta for

\mid x - \xi \mid &lt; \delta \Rightarrow \xi - \delta &lt; x &lt; \xi + \delta

but \xi is an upper bound for S, \xi &lt; \xi +\frac{\delta}{2}
and
\xi + \frac{\delta}{2} \in S since
f\left(\xi + \frac{\delta}{2}\right) &lt; \eta but \xi is an upper bound for S contradiction! so \eta - f(\xi) is not positive so still two choices less than or equal zero

Lets see if f(\xi) &gt; \eta
f is continuous at \xi , now pick \epsilon = f(\xi) - \eta
there exist delta

\mid f(x) - f(\xi) \mid &lt; f(\xi) - \eta for \mid x - \xi \mid &lt; \delta

-f(\xi) + \eta &lt; f(x) - f(\xi)

\eta &lt; f(x) for \xi - \delta &lt; x &lt; \xi + \delta

\xi - \delta is in S the definition of the upper bound which means f \left(\xi - \delta \right) should be less than \eta not larger contradiction so our epsilon is not positive

This idea from Wiki see this

The proof you proposed is completely different from the proof i wrote in my OP.

Sorry but i am not asking for a proof of the IVT but an explanation of the OP's proof